Using the Kunneth formula over Z, compute H_1(S^1 × S^1).
AZ
BZ ⊕ Z
CZ ⊕ Z ⊕ Z
D0
H_1(S^1 × S^1) = ⊕_{p+q=1} H_p(S^1) ⊗ H_q(S^1) (the Tor term vanishes since all homology groups of S^1 are free). The terms: H_0(S^1) ⊗ H_1(S^1) = Z ⊗ Z = Z, and H_1(S^1) ⊗ H_0(S^1) = Z ⊗ Z = Z. So H_1(T^2) = Z ⊕ Z, matching the known first homology of the torus. The two generators correspond to the two fundamental loops — one wrapping around each S^1 factor.
Question 2 True / False
The Kunneth formula with field coefficients gives H_n(X × Y; k) ≅ ⊕_{p+q=n} H_p(X; k) ⊗_k H_q(Y; k) with no correction term.
TTrue
FFalse
Answer: True
Over a field k, the Tor term vanishes because every module over a field is free (hence flat). The short exact Kunneth sequence becomes 0 → ⊕ H_p ⊗ H_q → H_n(X × Y) → 0, giving an isomorphism. This is why field coefficients are technically simpler: the Kunneth formula gives a clean tensor product decomposition. Over Z, torsion in the homology groups can produce Tor terms that contribute to the product homology.
Question 3 Multiple Choice
Let X = RP^2 (with H_0 = Z, H_1 = Z/2Z, H_2 = 0). Using Kunneth over Z, does H_2(RP^2 × RP^2) have torsion?
ANo — H_2 is free abelian
BYes — the Tor term Tor(H_1(RP^2), H_0(RP^2)) = Tor(Z/2Z, Z) = 0 contributes nothing, but Tor(H_0, H_1) also vanishes, so H_2 is the tensor sum only
CYes — Tor(Z/2Z, Z/2Z) = Z/2Z contributes a torsion summand
DCannot be determined without the full Kunneth computation
H_2(RP^2 × RP^2; Z): the tensor terms with p + q = 2 are H_0 ⊗ H_2 = 0, H_1 ⊗ H_1 = Z/2Z ⊗ Z/2Z = Z/2Z, H_2 ⊗ H_0 = 0. The Tor terms with p + (q-1) = 2, i.e., p + q = 2 in the Tor sum (shifted): Tor(H_1, H_0) = Tor(Z/2Z, Z) = 0, Tor(H_0, H_1) = Tor(Z, Z/2Z) = 0. Actually, the Tor contribution is ⊕_{p+q=n-1} Tor(H_p, H_q), so for n=2: Tor(H_0, H_0) + Tor(H_1, ...) — let me recompute. The Kunneth exact sequence is 0 → ⊕_{p+q=n} H_p ⊗ H_q → H_n → ⊕_{p+q=n-1} Tor(H_p, H_q) → 0. For n=2: tensor part = Z/2Z, Tor part (p+q=1): Tor(H_0, H_1) + Tor(H_1, H_0) = Tor(Z, Z/2Z) + Tor(Z/2Z, Z) = 0 + 0 = 0. So H_2 = Z/2Z, which IS torsion, but from the tensor term.
Question 4 Short Answer
Explain why the Kunneth formula is 'multiplicative' for Euler characteristics: χ(X × Y) = χ(X) · χ(Y).
Think about your answer, then reveal below.
Model answer: From the Kunneth formula (over a field, for simplicity): b_n(X × Y) = Σ_{p+q=n} b_p(X) · b_q(Y). Therefore χ(X × Y) = Σ_n (-1)^n b_n(X × Y) = Σ_n (-1)^n Σ_{p+q=n} b_p(X) · b_q(Y) = Σ_p Σ_q (-1)^{p+q} b_p(X) · b_q(Y) = (Σ_p (-1)^p b_p(X))(Σ_q (-1)^q b_q(Y)) = χ(X) · χ(Y). The key step is that (-1)^{p+q} = (-1)^p · (-1)^q, allowing the double sum to factor as a product.
This is one of the cleanest consequences of the Kunneth formula. It gives a quick way to compute Euler characteristics of products: χ(T^2) = χ(S^1) · χ(S^1) = 0 · 0 = 0, χ(S^2 × S^2) = 2 · 2 = 4, χ(S^1 × S^2) = 0 · 2 = 0. The multiplicativity fails for Betti numbers individually (b_1(T^2) = 2 ≠ b_1(S^1)^2 = 1), but works perfectly for the alternating sum.